Let $\mathrm{S}=\left\{\mathrm{A}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]: a, b, c, d \in\{0,1,2,3,4\}\right.$ and $\left.\mathrm{A}^2-4 \mathrm{~A}+3 \mathrm{I}=0\right\}$ be a set of $2 \times 2$ matrices. Then the number of matrices in S , for which the sum of the diagonal elements is equal to 4 , is :

Let $A=\left[\begin{array}{ccc}1 & 1 & 2 \\ -2 & 0 & 1 \\ 1 & 3 & 5\end{array}\right]$. Then the sum of all elements of the matrix $\operatorname{adj}\left(\operatorname{adj}\left(2(\operatorname{adj} \mathrm{~A})^{-1}\right)\right)$ is equal to:

If the system of equations

$x + 5y + 6z = 4$

$2x + 3y + 4z = 7$

$x + 6y + az = b$

has infinitely many solutions, then the point $(a, b)$ lies on the line

Let $\alpha, \beta \in \mathbb{R}$ be such that the system of linear equations

$ \begin{aligned} x + 2y + z &= 5 \\ 2x + y + \alpha z &= 5 \\ 8x + 4y + \beta z &= 18 \end{aligned} $

has no solution. Then $\frac{\beta}{\alpha}$ is equal to :